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I Rate of Change of Vector in Rotating Frame

  1. Dec 4, 2016 #1
    I recognize the rate of change of a vector in an inertial frame S can be related to the rate of change of the vector in a rotating frame S0 by the equation below taken from my textbook, where Ω is the angular velocity vector. $$\Big(\frac{dQ}{dt}\Big)_{S_{0}}= \Big(\frac{dQ}{dt}\Big)_{S} + \Omega \times Q$$
    However, I am unsure which frame of reference Q refers to in the cross product.
     
  2. jcsd
  3. Dec 4, 2016 #2
    Define Q.
     
  4. Dec 4, 2016 #3
    Q is an arbitrary vector such as a point of a ball in space.
     
  5. Dec 4, 2016 #4
    If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S). Your textbook may or may not have a picture illustrating this.
     
  6. Dec 4, 2016 #5
    The origins of both frames are coincident and Q can be described in either S or S0, but the textbook doesn't clarify which frame Q is described in for the cross product.
     
  7. Dec 5, 2016 #6

    vanhees71

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    Science Advisor
    2016 Award

    All equations are always in one frame or, even less confusing, you work with invariant objects, namely the vectors themselves, and not with their components. In this case, let's take ##\vec{e}_j## as the time-independent basis vectors in an inertial reference frame and ##\vec{e}_k'## the time-dependent vectors in the rotating frame. Then there's a rotation matrix ##D_{kj}## such that
    $$\vec{e}_k'=D_{kj} \vec{e}_j.$$
    Now take any vector ##\vec{A}##. You have
    $$\frac{\mathrm{d}}{\mathrm{d} t} \vec{A}=\dot{\vec{A}}=\frac{\mathrm{d}}{\mathrm{d} t} (A_k' \vec{e}_k').$$
    Now you have
    $$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + A_k' \dot{\vec{e}}_k'. \qquad (1)$$
    But now
    $$\dot{\vec{e}}_k'=\dot{D}_{kj} \vec{e}_j=\dot{D}_{kj} (D^{-1})_{jl} \vec{e}_l'=\epsilon_{klm} \Omega_m' \vec{e}_l', \qquad (2)$$
    where I have used that because of
    $$\hat{D} \hat{D}^T=\hat{1}$$
    the matrix ##\dot{D}_{kj} (D^{-1})_{jl}## is antisymmetric (prove it!). Plugging this in (2) in (1) finally gives
    $$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + \epsilon_{klm} A_k' \Omega_m' \vec{e}_l = (\dot{A}_l'+\epsilon_{lmk} \Omega_m' A_k') \vec{e}_l'.$$
    This means that for the components of ##\dot{\vec{A}}## you get
    $$\mathrm{D}_t \underline{A}'=\mathrm{d}_t \underline{A}' + \underline{\Omega}' \times \underline{A}',$$
    where the underlined symbols mean the component-column vectors wrt. the basis ##\vec{e}_k'##.
     
  8. Dec 5, 2016 #7

    Nugatory

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    Staff: Mentor

    Choose some fairly easy case, such as a unit vector along the x axis in the ##S_0## frame, and try it both ways.... See which interpretation of the formula gives you a sensible result.

    I'm not suggesting this because it's the best way of resolving the ambiguity in the textbook, but because the exercise will go a long ways towards building your intuition around how these transformations work.
     
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